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Question 62 of 68

Q.The value of 𝛼 if the angle between 𝑝⃗ = 2𝛼2𝑖̂ βˆ’ 3𝛼𝑗̂ + π‘˜Μ‚ and π‘žβƒ— = 𝑖̂ + 𝑗̂ + π›Όπ‘˜Μ‚ is obtuse, is
(A) 𝑅 βˆ’ [0, 1]
(B) (0, 1)
(C) [0, ∞)
(D) [1, ∞)

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The angle between two vectors is obtuse when their dot product is negative. Solving pβƒ—β‹…qβƒ—<0\vec{p} \cdot \vec{q} < 0 gives 2Ξ±2βˆ’2Ξ±<02\alpha^2 - 2\alpha < 0, which holds for α∈(0,1)\alpha \in (0, 1). The correct option is (B).

The key to this problem lies in a single, elegant fact: the sign of the dot product tells you everything about the angle between two vectors. If the dot product is positive, the angle is acute (less than 90Β°). If it's zero, the vectors are perpendicular. If it's negative, the angle is obtuse β€” greater than 90Β° and less than 180Β°.

So when a problem says "the angle is obtuse," it's handing you a direct inequality: p⃗⋅q⃗<0\vec{p} \cdot \vec{q} < 0. No need to compute the angle itself, no inverse cosine, no magnitudes — just the dot product and a simple inequality.

Let's apply this.

  1. Write down the vectors clearly.

    pβƒ—=2Ξ±2i^βˆ’3Ξ±j^+1k^\vec{p} = 2\alpha^2 \hat{i} - 3\alpha \hat{j} + 1 \hat{k}

    q⃗=1i^+1j^+αk^\vec{q} = 1 \hat{i} + 1 \hat{j} + \alpha \hat{k}

  2. Compute the dot product.

    Multiply corresponding components and add:

    pβƒ—β‹…qβƒ—=(2Ξ±2)(1)+(βˆ’3Ξ±)(1)+(1)(Ξ±)\vec{p} \cdot \vec{q} = (2\alpha^2)(1) + (-3\alpha)(1) + (1)(\alpha)

    =2Ξ±2βˆ’3Ξ±+Ξ±= 2\alpha^2 - 3\alpha + \alpha

    =2Ξ±2βˆ’2Ξ±= 2\alpha^2 - 2\alpha

  3. Set up the obtuse condition.

    For an obtuse angle, p⃗⋅q⃗<0\vec{p} \cdot \vec{q} < 0:

    2Ξ±2βˆ’2Ξ±<02\alpha^2 - 2\alpha < 0

  4. Solve the quadratic inequality.

    Factor out the common term:

    2Ξ±(Ξ±βˆ’1)<02\alpha(\alpha - 1) < 0

    Divide both sides by 2 (positive, so inequality direction stays the same):

    Ξ±(Ξ±βˆ’1)<0\alpha(\alpha - 1) < 0

    This is a simple quadratic inequality. The product of two numbers is negative when one is positive and the other is negative. The roots of Ξ±(Ξ±βˆ’1)=0\alpha(\alpha - 1) = 0 are Ξ±=0\alpha = 0 and Ξ±=1\alpha = 1.

    For a quadratic with a positive leading coefficient, the expression is negative between the roots.

    So Ξ±(Ξ±βˆ’1)<0\alpha(\alpha - 1) < 0 when 0<Ξ±<10 < \alpha < 1. …

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